Get subject-specific peak velocity and age at peak velocity values from linear mixed spline models

Question

I am fitting a linear mixed effects model with a natural spline function for age in order to describe the nonlinear trajectory for a repeated outcome y (bone mineral content in grams) across time (age in years).

How can I solve the spline derivatives to get the velocity curve and estimate for each individual their peak velocity (grams/year) and age at peak velocity (years) from this model?

This is the example data

dat <- structure(list(id = c(1001L, 1001L, 1001L, 1001L, 1001L, 1002L, 
    1003L, 1004L, 1004L, 1004L, 1004L, 1004L, 1004L, 1004L, 1005L, 
    1005L, 1005L, 1005L, 1005L, 1006L, 1006L, 1006L, 1006L, 1006L, 
    1007L, 1007L, 1008L, 1008L, 1008L, 1008L, 1008L, 1009L, 1009L, 
    1009L, 1010L, 1010L, 1010L, 1011L, 1012L, 1012L, 1012L, 1013L, 
    1013L, 1014L, 1015L, 1015L, 1015L, 1016L, 1016L, 1016L, 1016L, 
    1016L, 1017L, 1017L, 1018L, 1020L, 1020L, 1021L, 1021L, 1021L, 
    1021L, 1022L, 1022L, 1023L, 1023L, 1023L, 1023L, 1023L, 1023L, 
    1023L, 1023L, 1023L, 1023L, 1024L, 1024L, 1024L, 1024L, 1024L, 
    1025L, 1025L, 1025L, 1026L, 1026L, 1026L, 1026L, 1027L, 1027L, 
    1028L, 1028L, 1028L, 1028L, 1028L, 1028L, 1028L, 1029L, 1029L, 
    1029L, 1029L, 1029L, 1029L, 1030L, 1030L, 1030L, 1030L, 1030L, 
    1030L, 1030L, 1030L, 1031L, 1031L, 1031L, 1031L, 1032L, 1032L, 
    1032L, 1032L, 1032L, 1033L, 1033L, 1033L, 1033L, 1034L, 1034L, 
    1034L, 1034L, 1034L, 1035L, 1035L, 1036L, 1037L, 1037L, 1037L, 
    1037L, 1039L, 1039L, 1040L, 1040L, 1040L, 1040L, 1040L, 1040L, 
    1041L, 1041L, 1041L, 1041L, 1041L, 1041L, 1042L, 1042L, 1042L, 
    1042L, 1042L, 1042L, 1042L, 1043L, 1043L, 1043L, 1043L, 1044L, 
    1044L, 1044L, 1045L, 1045L, 1045L, 1045L, 1045L, 1045L, 1047L, 
    1048L, 1048L, 1049L, 1049L, 1049L, 1049L, 1051L, 1051L, 1052L, 
    1052L, 1052L, 1052L, 1052L, 1053L, 1053L, 1053L, 1053L, 1053L, 
    1054L, 1054L, 1054L, 1054L, 1054L, 1054L, 1054L, 1054L, 1056L, 
    1056L, 1056L, 1056L, 1057L, 1057L, 1058L, 1058L, 1058L, 1058L, 
    1058L, 1060L, 1060L, 1060L, 1061L, 1061L, 1061L, 1061L, 1061L, 
    1062L, 1062L, 1062L, 1062L, 1062L, 1063L, 1063L, 1063L, 1064L, 
    1064L, 1064L, 1064L, 1065L, 1065L, 1066L, 1066L, 1066L, 1066L, 
    1066L, 1066L, 1067L, 1067L, 1067L, 1068L, 1068L, 1068L, 1068L, 
    1068L, 1068L, 1068L, 1069L, 1070L, 1070L, 1070L, 1071L, 1071L, 
    1071L, 1072L, 1072L, 1072L, 1072L, 1072L, 1073L, 1073L, 1073L, 
    1073L, 1074L, 1074L, 1074L, 1075L, 1075L, 1075L, 1075L, 1075L, 
    1075L, 1076L, 1076L, 1076L, 1077L, 1077L, 1077L, 1077L, 1077L, 
    1077L, 1078L, 1078L, 1078L, 1078L, 1078L, 1078L, 1078L, 1080L, 
    1080L, 1080L, 1080L, 1081L, 1081L, 1082L, 1082L, 1082L, 1083L, 
    1083L, 1084L, 1085L, 1085L, 1085L, 1085L, 1085L, 1085L, 1086L, 
    1086L, 1086L, 1087L, 1087L, 1087L, 1087L, 1087L, 1087L, 1087L, 
    1087L, 1088L, 1088L, 1088L, 1088L, 1089L, 1089L, 1089L, 1089L, 
    1089L, 1090L, 1090L, 1091L, 1091L, 1091L, 1091L, 1091L, 1092L, 
    1092L, 1092L, 1092L, 1092L, 1093L, 1093L, 1093L, 1093L, 1094L, 
    1094L, 1094L, 1094L, 1094L, 1095L, 1095L, 1095L, 1095L, 1096L, 
    1097L, 1097L, 1098L, 1098L, 1098L, 1098L, 1098L, 1099L, 1099L, 
    1099L, 1099L, 1099L, 1099L, 1099L, 1099L, 1100L, 1100L, 1100L, 
    1101L, 1101L, 1101L, 1101L, 1103L, 1103L, 1103L, 1103L, 1103L, 
    1103L, 1103L, 1104L, 1104L, 1104L, 1104L, 1105L, 1105L, 1105L, 
    1106L, 1106L, 1106L, 1106L, 1106L, 1106L, 1106L, 1106L, 1106L, 
    1107L, 1108L, 1110L, 1111L, 1112L, 1117L, 1123L), y = c(1934.047646, 
    1075.598345, 1956.214821, 2000.38538, 2000.38538, 732.315937, 
    3119.86, 624.951231, 791.2764892, 1884.530826, 624.951231, 1047.57, 
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    2547.870529, 2467.385, 1181.635212, 1181.635212, 565.306282, 
    2016.027874, 2016.027874, 712.6134567, 635.2537841, 2167.362267, 
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This is the code for fitting the model and plotting the mean predicted trajectory

    library(nlme)
    library(splines)
    library(tidyverse)

# LINEAR MIXED MODEL WITH NATURAL SPLINE FUNCTION FOR AGE
    nspline_model <- lme(y ~ ns(age, df = 4), data = dat, random = ~ age | id)
    
# PLOT MEAN FITTED NATURAL SPLINE TRAJECTORY
    dat$ns_pred <- fitted(nspline_model, level = 0)
    ggplot(data = dat, aes(x = age, y = ns_pred)) + geom_line() 

This is a possible solution using the getPeakTrough function and smooth.spline but i am not sure if it is correct - or how to get subject specific peak velocities rather than just the mean values.

    # POSSIBLE SOLUTION FOR THE NATURAL SPLINE MODEL???
    x <- dat$age
    y <- fitted(nspline_model, level = 0)
    y <- predict(smooth.spline(x, y, df = 4), x, deriv = 1)$y
    sitar::getPeakTrough(x, y, peak = T)
    # mean age at peak velocity = 11.82447 years; 
    # mean peak velocity 187.13404 grams/year

Answer 1

Note: My answer here also serves a tutorial/vignette for SplinesUtils.

---

Package SplinesUtils (https://github.com/ZheyuanLi/SplinesUtils) is now enhanced to support linear mixed models fitted by nlme::lme. (The previous version is motivated by this SO thread: Export fitted regression splines (constructed by 'bs' or 'ns') as piecewise polynomials).

library(SplinesUtils)

Model description

nspline_model <- lme(y ~ ns(age, df = 4), data = dat, random = ~ age | id)

This is a linear mixed effect model with:

• a fixed effect ns(age, df = 4), which is a natural cubic spline of age:

• a random effect ~ age | id, which is a linear function of age for each id.

The fixed effect reflects the average relationship between y and age over the population; while the random effect is the sample-specific deviation from this overall relationship.

Use SplinesUtils to process the overall relationship

spl_population <- RegSplineAsPiecePoly(nspline_model, "ns(age)")
#Error: 
#  Required SplineTerm not found! Available terms are:
#    * ns(age, df = 4)

In above, I purposely wrote a line of wrong code. This error has confused some users (Finding coefficient of a spline regression in a multiple regression in R). The issue is that users have to correctly provide the name of the spline that is internally known to the model fitting function. From version 0.2, the error message is better formatted, listing such internal names one per line. Basically, users just need to copy the right one and then paste it to RegSplineAsPiecePoly:

spl_population <- RegSplineAsPiecePoly(nspline_model, "ns(age, df = 4)")

By now the reparametrization is complete. There are various ways to inspect the exported piecewise polynomials.

The most fundamental one is to get coefficients in a matrix form:

spl_population$PiecePoly$coef
#             [,1]       [,2]        [,3]          [,4]
#[1,] 5.684342e-13 645.727356 1383.877112  1.871766e+03
#[2,] 9.419308e+01 220.369323  209.951537 -3.223616e-01
#[3,] 1.782352e-12  26.623843  -30.064255 -4.796068e-01
#[4,] 1.872590e+00  -6.240304    1.432464  9.595289e-03

To get a better idea of the meaning of these numbers, we can print out the formatted strings of polynomial equations:

spl_population  ## or print(spl_polulation)
#4 piecewise polynomials of degree 3 are constructed!
#Use 'summary' to export all of them.
#The first 4 are printed below.
#5.68e-13 + 94.2 * (x - 7.99) + 1.78e-12 * (x - 7.99) ^ 2 + 1.87 * (x - 7.99) ^ 3
#646 + 220 * (x - 12.7) + 26.6 * (x - 12.7) ^ 2 - 6.24 * (x - 12.7) ^ 3
#1.38e+03 + 210 * (x - 15.8) - 30.1 * (x - 15.8) ^ 2 + 1.43 * (x - 15.8) ^ 3
#1.87e+03 - 0.322 * (x - 22.6) - 0.48 * (x - 22.6) ^ 2 + 0.0096 * (x - 22.6) ^ 3

Formatted strings always use "x" as variable name. In this context, we know that "x" represents age.

Note that the print method written for a "PiecePoly" object is to print out formatted polynomial equations. To see what spl_polulation actually is, use:

str(spl_population)
#List of 2
# $ PiecePoly:List of 2
#  ..$ coef : num [1:4, 1:4] 5.68e-13 9.42e+01 1.78e-12 1.87 6.46e+02 ...
#  ..$ shift: logi TRUE
# $ knots    : num [1:5] 7.99 12.73 15.76 22.64 39.3
# - attr(*, "class")= chr [1:2] "PiecePoly" "RegSpline"

There is actually an intermediate step from this raw form to print, which is summary. The summary function constructs formatted strings using the raw coefficients. Let's verify this:

summary(spl_population)
#4 piecewise polynomials of degree 3 are constructed!

You did not see the equations because by default, they are hidden (because I want to avoid printing out all pieces to flush your screen). You can try:

str(summary(spl_population))
#4 piecewise polynomials of degree 3 are constructed!
#List of 4
# $ : chr "5.68e-13 + 94.2 * (x - 7.99) + 1.78e-12 * (x - 7.99) ^ 2 + 1.87 * (x - 7.99) ^ 3"
# $ : chr "646 + 220 * (x - 12.7) + 26.6 * (x - 12.7) ^ 2 - 6.24 * (x - 12.7) ^ 3"
# $ : chr "1.38e+03 + 210 * (x - 15.8) - 30.1 * (x - 15.8) ^ 2 + 1.43 * (x - 15.8) ^ 3"
# $ : chr "1.87e+03 - 0.322 * (x - 22.6) - 0.48 * (x - 22.6) ^ 2 + 0.0096 * (x - 22.6) ^ 3"

If users wonder why these equations are polynomials of (x - some_number) rather than x, consult ?PiecePoly for the difference between "shifted form" and "unshifted form".

Plotting this spline is rather straightforward:

plot(spl_population)

In case you find the the plot is not visually smooth enough, make the parameter spread bigger (default value is 3; grow it slowly!):

plot(spl_population, spread = 5)

The plot function can also sketch the 1st (or any) derivative of the spline, by using argument deriv. Try

plot(spl_population, spread = 5, deriv = 1)

To find all local extrema of the spline, use solve + predict:

( xe <- solve(spl_population, b = 0, deriv = 1) )
#[1] 22.45925
( ye <- predict(spl_population, xe) )
#[1] 1871.8
plot(spl_population)
points(xe, ye)

To find all local extrema of the 1st derivative of the spline (which is what you are looking for), we need to take an extra derivative:

( xa <- solve(spl_population, b = 0, deriv = 2) )
#[1] 14.15315
( ya <- predict(spl_population, xa, deriv = 1) )
#[1] 258.2323
plot(spl_population, deriv = 1)
points(xa, ya)

The result you obtained from sitar::getPeakTrough in your question is close, but not correct. Because by using smooth.spline you do some approximation, while the result obtained by SplinesUtils is accurate.

Use SplinesUtils to process the sample-specific relationship

By adding the sample-specific random linear function of age to the population spline, we get the sample-specific spline.

The random line can be obtained as the fitted coefficients of the random effect:

( random_line <- random.effects(nspline_model) )
#     (Intercept)         age
#1001  105.366791 -17.3226305
#1002    2.118663  -3.1252693
#1003 -124.348097  25.3333794
#1004   35.118293  -9.2338755
#...          ...         ...

At this stage it is more convenient to export the population spline with "unshifted form":

spl_population_unshifted <- RegSplineAsPiecePoly(nspline_model, "ns(age, df = 4)", FALSE)
spl_population_unshifted
#4 piecewise polynomials of degree 3 are constructed!
#Use 'summary' to export all of them.
#The first 4 are printed below.
#-1.71e+03 + 453 * x - 44.9 * x ^ 2 + 1.87 * x ^ 3
#1.5e+04 - 3.49e+03 * x + 265 * x ^ 2 - 6.24 * x ^ 3
#-1.5e+04 + 2.22e+03 * x - 97.8 * x ^ 2 + 1.43 * x ^ 3
#1.52e+03 + 36.2 * x - 1.13 * x ^ 2 + 0.0096 * x ^ 3

Now for example, to get the sample-specific spline for id = 1001, we can update the polynomial coefficients:

( coef_population <- spl_population_unshifted$PiecePoly$coef )
#            [,1]         [,2]          [,3]          [,4]
#[1,] -1708.58691 15031.740239 -14997.457442  1.521760e+03
#[2,]   452.99242 -3491.784683   2224.770784  3.615668e+01
#[3,]   -44.89601   264.959868    -97.787157 -1.131417e+00
#[4,]     1.87259    -6.240303      1.432464  9.595289e-03

coef_1001 <- coef_population  ## initialize
coef_1001[1, ] <- coef_1001[1, ] + random_line[1, 1]  ## update intercept
coef_1001[2, ] <- coef_1001[2, ] + random_line[1, 2]  ## update slope

The coefficient matrix alone is not a "PiecePoly" object, so we can't plot it nor solve it. But we can do some "hacking":

spl_1001_unshifted <- spl_population_unshifted
spl_1001_unshifted$PiecePoly$coef <- coef_1001

Now we can plot it, find and overlay its extrema:

plot(spl_1001_unshifted)
( xe_1001 <- solve(spl_1001_unshifted, b = 0, deriv = 1) )
#[1] 20.72561
if (length(xe_1001)) {
  ( ye_1001 <- predict(spl_1001_unshifted, xe_1001) )
  #[1] 1606.861
  points(xe_1001, ye_1001)
}

(Note: The if wrapper is generally necessary because xe_1001 can be numeric(0), which means that the spline is monotone and there is no extremum!)

We can also plot its 1st derivative then overlays its extrema (which is what you are looking for):

plot(spl_1001_unshifted, deriv = 1)
( xa_1001 <- solve(spl_1001_unshifted, b = 0, deriv = 2) )
#[1] 7.991786 14.153151 39.304586
if (length(xa_1001)) {
  ( ya_1001 <- predict(spl_1001_unshifted, xa_1001, deriv = 1) )
  #[1] 76.87045 240.90965 -25.63581
  points(xa_1001, ya_1001)
}

This plot shows that the two boundaries are also marked. This is because natural cubic spline are constrained to have zero 1st derivative at two ends. So no surprise. Of course, the computer program, due to small rounding error, may or may not include these two ends. For example, They are not caught in the population case (see the last figure in the previous section). But If they do show up, just exclude them and keep the middle ones.

To conclude, as soon as you learn how to do this for one sample, you can write a loop to process all samples.

In reply to your comment

The only challenge is how to get standard errors/confidence intervals for the mean age at peak velocity (i.e. to estimate the values for every person).

That is another question well worth researching, whose answer needs equally (maybe more) endeavor as this one. The Stack Overflow policy may expect you to ask a new question for that, otherwise our discussion here will never end.

Answer 2

One way to address this is to compute the slopes between each sequential point for each individual. You can then take the age of the upper point for the period with the maximum slope. The code below achieves this. Note you can substitute ns_pred for ls_pred in the mutate step to switch models.

data2 <- data %>%
  group_by(id) %>%
  filter(n() > 2) %>%
  arrange(age) %>%
  mutate(slope = (ns_pred - lag(ns_pred)) / (age - lag(age))) %>%
  arrange(slope) %>%
  top_n(n = 1)

It seems you have some sparse entries so I also filtered out individuals with 2 or fewer observations. Also, because it seems there are some individuals with long gaps between observations, perhaps it would be more appropriate to average the two ages which flank the maximum slope period. However this is more of a discretionary questions for the researcher.